To what temperature will 7630 J of work raise 2.9 kg of water initially at 9.8°C?

Answer 1

Use dQ = m c dT where m = mass of water, dT = change in temperature and c = specific heat of water

dQ = 7630 J, dT = Tf-9.8C, c = 4.186 J/gm/C, and m = 2.9x10^3 gm


Then dQ = m c dT = mcTf - 9.8 mc

dQ/mc = Tf -9.8

Tf = dQ/mc +9.8 = 10.43 C

Answer 2

Q= m*c*T with :
m = mass of water = 2.9 kg = 2900 g
T = change in temperature
and c = specific heat of water = 4186 J/kg = 4.186 J/g
Q = 7630 J

So T= 7630/(2900*4.186)=0.62853188790220274478145542613309

final temperature is 9.8+T#10.42853

Answer 3

Q = mcT, where T is the change of temperature.

Q = W = 7630 J
specific heat capacity for water, c = 4200 Jkg-1K-1
T = X - 9.8, where X is the final temperature in °C.

So,
7630 = (2.9)(4200)(X - 9.8)
X - 9.8 = 0.6264
X = 10.426 °C

Answer 4

we use joules law. w=jh
here j=1(si system and so j=1)
so h=7630 j
h=mst
s=4200,m=2.9.
7630=2.9*4200*(t-9.8)
t=10.4264